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The term vanishes because differentiability ensures that both $\Delta f$ and $\Delta g$ are proportional to the input increment $h$ (i.e., of order $O(h)$). When divided by $h$, the product $\frac{\Delta f \Delta g}{h}$ becomes an expression of order $O(h^2)/h = O(h)$, which approaches zero as $h \to 0$.

Conditions: Functions $f$ and $g$ are differentiable at the point of interest; Input increment $h$ approaches zero

The term vanishes because differentiability ensures that both $\Delta f$ and $\Delta g$ are proportional to the input increment $h$ (i.e., of order $O(h)$). When divided by $h$, the product $\frac{\Delta f \Delta g}{h}$ becomes an expression of order $O(h^2)/h = O(h)$, which approaches zero as $h \to 0$.

Conditions: Functions $f$ and $g$ are differentiable at the point of interest; Input increment $h$ approaches zero